Calderon s inverse conductivity problem in the plane

Transcription

1 Calderon s inverse conductivity problem in the plane Kari Astala and Lassi Päivärinta Abstract We show that the Dirichlet to Neumann map for the equation σ u = 0 in a two dimensional domain uniquely determines the bounded measurable conductivity σ. This gives a positive answer to a question of A. P. Calderón from Earlier the result has been shown only for conductivities that are sufficiently smooth. Contents 1. Introduction and outline of the method 2. Beltrami equation and Hilbert transform 3. Beltrami operators 4. Complex geometric optics solutions 5. k -equations 6. From Λ σ to τ 7. Subexponential growth 8. Transport matrix The research of both authors is supported by the Academy of Finland 1

2 1 Introduction and outline of the method In 1980 A. P. Calderón [9] posed the following problem: Suppose that Ω R n is a bounded domain with connected complement and σ : Ω (0, ) is measurable and bounded away from zero and infinity. Let u H 1 (Ω) be the unique solution to (1.1) (1.2) σ u = 0 in Ω, u Ω = φ H 1/2 ( Ω). The inverse conductivity problem of Calderón is then to recover σ from the boundary measurements, from the Dirichlet to Neumann map Λ σ : φ σ u. ν Ω Here ν is the unit outer normal to the boundary and the derivative σ u/ ν exists as an element of H 1/2 ( Ω), defined by (1.3) σ u ν, ψ = σ u ψ dm, Ω where ψ H 1 (Ω) and dm denotes the Lebesgue measure. The aim of this paper is to give a positive answer to Calderón s question in dimension two. More precisely, we prove Theorem 1 Let Ω R 2 be a bounded simply connected domain and σ i L (Ω), i = 1, 2. Suppose that there is a constant c > 0 such that c 1 σ i c. If Λ σ1 = Λ σ2 then σ 1 = σ 2. Note, in particular, that no regularity is required for the boundary. Our approach to Theorem 1 yields, in principle, also a method to construct σ from the Dirichlet to Neumann operator Λ σ. For this see Section 8. The inverse problem to determine σ from Λ σ is also known as Electrical Impedance Tomography. It has been proposed as a valuable diagnostic tool especially for detecting breast cancer [10]. A review for medical applications is given in [11]. For statistical methods in electrical impedance tomography see [17]. That Λ σ uniquely determines σ was established in dimension three and higher for smooth conductivities by J. Sylvester and G. Uhlmann [26] in

3 In dimension two A. Nachman [19] produced in 1995 a uniqueness result for conductivities with two derivatives. For piecewise analytic conductivities the problem was solved by Kohn and Vogelius [15], [16]. The regularity assumptions have since been relaxed by several authors (cf. [20], [21]) but the original problem of Calderón has still remained unsolved. The largest class of potentials where the uniqueness has been shown so far is W 3/2, (Ω) in dimensions three and higher [23] and W 1,p (Ω), p > 2, in dimension two [8]. The original approach in [26] and [19] was to reduce the conductivity equation (1.1) to the Schrödinger equation by substituting v = σ 1/2 u. Indeed, after such a substitution v satisfies v qv = 0 where q = σ 1/2 σ 1/2. This explains why in this method one needs two derivatives. For the numerical implementation of [19] see [24]. Following the ideas of Beals and Coifman [6], Brown and Uhlman [8] found a first order elliptic system equivalent to (1.1). Indeed, by denoting ( ) ( ) v u = σ 1/2 w u one obtains the system where D = D ( ) ( ) v v = Q, w w ) ( 0, Q = 0 ( 0 ) q q 0 and q = 1 log σ. This allowed Brown and Uhlman to work with conductivities with only one derivative. Note however, that the assumption 2 σ W 1,p (Ω), p > 2, made in [8] implies that σ is Hölder continuous. From the viewpoint of applications this is still not satisfactory. Our starting point is to replace (1.1) with an elliptic equation that does not require any differentiability of σ. We will base our argument to the fact that if u H 1 (Ω) is a real solution of (1.1) then there exists a real function v H 1 (Ω), called the σ-harmonic conjugate of u, such that f = u + iv satisfies the R-linear Beltrami equation (1.4) f = µ f, where µ = (1 σ)/(1 + σ). In particular, note that µ is real valued. The assumptions for σ imply that µ L κ < 1, and the symbol κ will retain this role throughout the paper. 3

4 The structure of the paper is the following: Since the σ-harmonic conjugate is unique up to a constant we can define the µ-hilbert transform H µ : H 1/2 ( Ω) H 1/2 ( Ω) by H µ : u Ω v Ω. We show in Section 2 that the Dirichlet to Neumann map Λ σ uniquely determines H µ and vice versa. Theorem 1 now implies the surprising fact that H µ uniquely determines µ in equation (1.4) in the whole domain Ω. Recall that a function f Hloc 1 (Ω) satisfying (1.4) is called a quasireqular mapping; if it is also a homeomorphism then it is called quasiconformal. These have a well established theory, cf. [2], [5], [12], [18], that we will employ at several points in the paper. The Hloc 1 -solutions f to (1.4) are automatically continuous and admit a factorization f = ψ H, where ψ is C-analytic and H is a quasiconformal homeomorphism. Solutions with less regularity may not share these properties [12]. The basic tools to deal with the Beltrami equation are two linear operators, the Cauchy transform P = 1 and the Beurling transform S = 1. In Section 3 we recall the basic properties of these operators with some useful preliminary results. It is not difficult to see, c.f. Section 2, that we can assume Ω = D, the unit disk of C, and that outside Ω we can set σ 1, i.e., µ 0. In Section 4 we establish the existence of the geometric optics solution of (1.4) that have the form (1.5) f µ (z, k) = e ikz M µ (z, k), where (1.6) M µ (z, k) = 1 + O ( ) 1 z as z. As in the smooth case these solutions obey a -equation also in the k variable. However, their asymptotics as k are now more subtle and considerably more difficult to handle. It turns out that it is instructive to consider the conductivities σ and σ 1, or equivalently the Beltrami coefficients µ and µ, simultaneously. By defining (1.7) h + = 1 2 (f µ + f µ ), h = i 2 (f µ f µ ) we show in Section 5 that with respect to the variable k, h + and h satisfy the equations (1.8) k h + = τ µ h, k h = τ µ h + 4

5 where the scattering coefficient τ µ = τ µ (k) is defined by (1.9) τ µ (k) = 1 ( ) z Mµ M µ d z dz. 4πi The remarkable fact in (1.8) is that the coefficient τ µ (k) does not depend on the space variable z; the idea to use such a phenomenon is due to Beals and Coifman [6]. In Section 6 we show that Λ σ uniquely determines the scattering coefficient τ µ (k) as well as the geometric optics solutions f µ and f µ outside D. The crucial problem in the proof of Theorem 1 is the behavior of the function M µ (z, k) 1 = e ikz f µ (z, k) 1 with respect to the k-variable. In the case of [19] and [8] the behaviour is roughly like k 1. In the L -case we cannot expect such a good behavior. Instead, we can show that M µ (z, k) grows at most subexponentially in k. This is the key tool to our argument and it takes a considerable effort to prove this. Precisely, we show in Section 7 that f µ (z, k) = exp(ikϕ(z, k)) where ϕ is a quasiconformal homeomorphism in the z-variable and satisfies the nonlinear Beltrami equation (1.10) z ϕ = k k µ(z)e k(ϕ(z)) z ϕ with the boundary condition (1.11) ϕ(z) = z + O ( ) 1 z at infinity. Here the unimodular function e k is given by (1.12) e k (z) = e i(kz+kz). The main result in Section 7 is that the unique solution of (1.10) and (1.11) obeys the property that (1.13) ϕ(z, k) z 0 as k, uniformly in z. Section 8 is devoted to the proof of Theorem 1. Since µ = f µ / fµ and f for a non-constant quasiregular map f can vanish only in a set of Lebesgue measure zero, we are reduced to determine the function f µ in the 5

6 interior of D. As said before, we already know these functions outside of D. The key ingredient to solve this problem is the so-called transport matrix that transforms the solutions outside D to solutions inside. We show that this matrix is uniquely determined by Λ σ. At this point one may work either with equation (1.1) or equation (1.4). We chose to go back to the conductivity equation since it slightly simplifies the formulas. More precisely, we set (1.14) u 1 = h + ih and u 2 = i(h + + ih ). Then u 1 and u 2 are complex solutions of the conductivity equations (1.15) σ u 1 = 0 and 1 σ u 2 = 0, respectively, and of the k -equation (1.16) k u j = i τ µ (k)u j, j = 1, 2, with the asymptotics u 1 = e ikz (1 + O(1/z)) and u 2 = e ikz (i + O(1/z)) in the z-variable. Uniqueness of (1.15) with these asymptotics gives that in the smooth case u 1 is exactly the exponentially growing solution of [19]. We then choose a point z 0 C, z 0 > 1. It is possible to write for each z, k C (1.17) u 1 (z, k) = a 1 u 1 (z 0, k) + a 2 u 2 (z 0, k) u 2 (z, k) = b 1 u 1 (z 0, k) + b 2 u 2 (z 0, k) where a j = a j (z, z 0 ; k) and b j = b j (z, z 0 ; k) are real valued. The transport matrix Tz,z σ 0 (k) is now defined by ( ) (1.18) Tz,z σ a1 a 0 (k) = 2. b 1 b 2 It is an invertible 2 2 real matrix depending on z, z 0 and k. The proof of Theorem 1 is thus reduced to Theorem 2 Assume that Λ σ = Λ σ for two L -conductivities σ and σ. Then for all z, k C and z 0 > 1 the corresponding transport matrices T σ z,z 0 (k) and T σ z,z 0 (k) are equal. The idea behind the proof is to use the Beals-Coifman method in an efficient manner and to show that the functions (1.19) α(k) = a 1 (k) + ia 2 (k) and β(k) = b 1 (k) + ib 2 (k) 6

7 both satisfy with respect to the parameter k the Beltrami equation (1.20) k α = ν z0 (k) k α. Here the coefficient (1.21) ν z0 (k) = i h (z 0, k) h + (z 0, k) is determined by the data as proved in Section 6. Moreover it satisfies ν z0 (k) q < 1, where the number q is independent of k (or z). These facts and the subexponential growth of the solutions serve as the key elements for the proof of Theorem 2. 2 Beltrami equation and Hilbert transform In a general domain Ω we identify H 1/2 ( Ω) = H 1 (Ω)/H0(Ω). 1 When Ω has enough regularity, trace theorems and extension theorems [27] readily yield the standard interpretation of H 1/2 ( Ω). The Dirichlet condition (1.2) is consequently defined in the Sobolev sense, requiring that u φ H0(Ω) 1 for the element φ H 1/2 ( Ω). Furthermore, H 1/2 ( Ω) = H 1/2 ( Ω) and via (1.3) it is then clear that Λ σ becomes a well-defined and bounded operator from H 1/2 ( Ω) to H 1/2 ( Ω). In this setup Theorem 1 quickly reduces to the case where the domain Ω is the unit disk. In fact, let Ω be a simply connected domain with Ω D and let σ and σ be two L -conductivities on Ω with Λ σ = Λ σ. Continue both conductivities as the constant 1 outside Ω to obtain new L -conductivities σ 0 and σ 0. Given φ H 1/2 ( D), let u 0 H 1 (D) be the solution to the Dirichlet problem σ 0 u 0 = 0 in D, u 0 D = φ. Assume also that ũ H 1 (Ω) is the solution to σ ũ = 0 in Ω, ũ u 0 H0(Ω). 1 Then ũ 0 = ũχ Ω + u 0 χ D\Ω H 1 (D) since zero extensions of H0(Ω) 1 functions remain in H 1. Moreover, an application of the definition (1.3) with the condition Λ σ = Λ σ yields that ũ 0 satisfies σ 0 ũ 0 = 0 in the weak sense. Since in D \ Ω we have u 0 ũ 0 and σ 0 σ 0, we obtain Λ σ0 φ = Λ σ0 φ, and this holds for all φ H 1/2 ( D). Thus if Theorem 1 holds for D we get σ 0 = σ 0 and especially that σ = σ. 7

8 From this on we assume that Ω = D, the unit disc in C. Let us then consider the complex analytic interpretation of (1.1). We will use the notations = 1 2 ( x i y ) and = 1 2 ( x + i y ); when clarity requires we may write = z or = z. For derivatives with respect to the parameter k we always use the notations k and k. We start with a simple lemma Lemma 2.1 Assume u H 1 (D) is real valued and satisfies the conductivity equation (1.1). Then there exists a function v H 1 (D), unique up to a constant, such that f = u + iv satisfies the R-linear Beltrami equation (2.1) f = µ f, where µ = (1 σ)/(1 + σ). Conversely, if f H 1 (D) satisfies (2.1) with a R-valued µ, then u = Re f and v = Im f satisfy (2.2) σ u = 0 and 1 v = 0, σ respectively, where σ = (1 µ)/(1 + µ). Proof: Denote by w the vectorfield w = ( σ 2 u, σ 1 u) where 1 = / x and 2 = / y for z = x + iy C. Then by (1.1) the integrability condition 2 w 1 = 1 w 2 holds for the distributional derivatives. Therefore there exists v H 1 (D), unique up to a constant, such that (2.3) (2.4) 1 v = σ 2 u 2 v = σ 1 u. It is a simple calculation to see that this is equivalent to (2.1). We want to stress that every solution of (2.1) is also a solution of the standard C-linear Beltrami equation (2.5) f = µ f but with a different, C-valued µ having though the same modulus as the old one. However, the uniqueness properties of (2.1) and (2.5) are quite different (cf. [28], [5]). Note also that the conditions for σ given in Theorem 1 imply the existence of a constant 0 κ < 1 such that µ(z) κ 8

9 holds for almost every z C. Since the function v in Lemma 2.1 is defined only up to a constant we will normalize it by assuming (2.6) v ds = 0. D This way we obtain a unique map H µ : H 1/2 ( D) H 1/2 ( D) by setting (2.7) H µ : u D v D. The function v satisfying (2.3), (2.4) and (2.6) is called the σ-harmonic conjugate of u and H µ the Hilbert transform corresponding to equation (2.1). Since v is the real part of the function g = if satisfying g = µ g, we have (2.8) H µ H µ u = H µ H µ u = u + u ds where D u ds = 1 2π D u ds. So far we have only defined H µ (u) for real-valued u. By setting H µ (iu) = ih µ (u) we have extended the definition of H µ (g) R-linearly to all C-valued g H 1/2 ( D). We also define Q µ : H 1/2 ( D) H 1/2 ( D) by (2.9) Q µ = 1 2 (I ih µ). Then g Q µ (g) + 1 g ds is a projection in 2 D H1/2 ( D). In fact (2.10) Q 2 µ(g) = Q µ (g) 1 4 g ds. The proof of the following lemma is straight forward. Lemma 2.2 If g H 1/2 ( D), the following conditions are equivalent, a) g = f D, where f H 1 (D) and satisfies (2.1). b) Q µ (g) is a constant. D D 9

10 We close this section by Proposition 2.3 The Dirichlet to Neumann map Λ σ uniquely determines H µ, H µ and Λ σ 1. Proof: Choose the counter clock-wise orientation for D and denote by T the tangential (distributional) derivative on D corresponding to this orientation. We will show that (2.11) T H µ (u) = Λ σ (u) holds in the weak sense. This will be enough since by (2.8) H µ uniquely determines H µ. Note also that µ = (1 σ 1 )/(1 + σ 1 ) and so Λ σ 1(u) = T H µ (u). By the definition of Λ σ we have ϕλ σ u ds = ϕ σ u dm, ϕ C (D). D D Thus, by (2.3), (2.4) and integration by parts, we get ϕλ σ u = ( 1 ϕ 2 v 2 ϕ 1 v) dm D D = D v T ϕ ds and (2.11) follows. 3 Beltrami operators The Beltrami differential equation (1.4) and its solutions are effectively governed and controlled by two basic linear operators, the Cauchy transform and the Beurling transform. Any analysis of (1.4) requires basic facts of these operators. We briefly recall those in this section. The Cauchy transform (3.1) P g(z) = 1 π C g(ω) ω z dm(ω) 10

11 acts as the inverse operator to ; P g = P g = g, for g C 0 (C). We recall some mapping properties of P in appropriate Lebesgue, Sobolev and Lipschitz spaces. Below we denote L p (Ω) = { g L p (C) g } C\Ω 0. Proposition 3.1 Let Ω C be a bounded domain and let 1 < q < 2 and 2 < p <. Then (i) P : L p (C) Lip α (C), where α = 1 2/p; (ii) P : L p (Ω) W 1,p (C) is bounded; (iii) P : L p (Ω) L p (C) is compact; (iv) P : L p (C) L q (C) C 0 (C) is bounded, where C 0 is the closure of C 0 in L. For proof of Proposition 3.1 we refer to [28], but see also [19]. The Beurling transform is formally determined by Sg = P g and more precisely as a principal value integral (3.2) Sg(z) = 1 π C g(ω) (ω z) 2 dm(ω). It is a Calderón-Zygmund operator with a holomorphic kernel. Since S is a Fourier multiplier operator with symbol (3.3) m(ξ) = ξ ξ, ξ = ξ 1 + iξ 2 we see, in particular, that S transforms the -derivatives to -derivatives, (3.4) S( ϕ) = ϕ, for ϕ S (C). Moreover, we have S = R iR 1 R 2 R 2 2, where R i s denote the Riesz-transforms. Also, it follows (cf. [25]) that (3.5) S : L p (C) L p (C), 1 < p <, and lim p 2 S L p L p = S L 2 L 2 = 1. It is because of (3.4) that the mapping properties of the Beurling transform control the solutions to the Beltrami equation (1.4). For instance, if 11

12 supp(µ) is compact as it is in our case, finding a solution to (1.4) with asymptotics ( ) 1 (3.6) f(z) = λz + O, z z is equivalent to solving and setting g = µsg + λµ f(z) = λz + P g(z), where P is the Cauchy transform. Therefore, if we denote by S the R-linear operator S(g) = S(g), we need to understand the mapping properties of P and the invertibility of the operator I µs in appropriate L p -spaces in order to determine to which L p -class the gradient of the solution to (1.4) belongs. Recently, Astala, Iwaniec and Saksman established through the fundamental theory of quasiconformal mappings the precise L p -invertibility range of these operators. Theorem 3.2 Let µ 1 and µ 2 be two C-valued measurable functions such that (3.7) µ 1 (z) + µ 2 (z) κ holds for almost every z C with a constant 0 κ < κ < p < 1 + 1/κ. Then the Beltrami operator Suppose that (3.8) B = I µ 1 S µ 2 S is bounded and invertible in L p (C), with norms of B and B 1 bounded by constants depending only on κ and p. Moreover, the bound in p is sharp; for each p 1 + κ and for each p 1 + 1/κ there are µ 1 and µ 2 as above such that B is not invertible in L p (C). For the proof see [4]. Since S L 2 L2 = 1, all operators in (3.8) are invertible in L 2 (C) as long as κ < 1. Thus Theorem 3.2 determines the interval around the exponent p = 2 where the invertibility remains true. Note that it is a famous open problem [14] whether it holds } S L p L {p p = max 1 1,. p 1 12

13 If this turns out to be the case, then µs L p L p µ L S L p L p < 1 whenever p < 1 + 1/ µ L. This would then give an alternative proof of Theorem 3.2. Theorem 3.2 has also nonlinear counterparts [4] yielding solutions to nonlinear uniformly elliptic PDE s; here see also [13], [5]. On the other hand, in two dimensions the uniqueness of solutions to general nonlinear elliptic systems is typically reduced to the study of the pseudoanalytic functions of Bers (cf [7], [28]). In the sequel we will need the following version of this principle. Proposition 3.3 Let F W 1,p loc (C) and γ Lp loc (C) for some p > 2. Suppose that for some constant 0 κ < 1, (3.9) F (z) κ F (z) + γ(z) F (z) holds for almost every z C. Then we have a) If F (z) 0 as z and γ has a compact support then F (z) 0. b) If for large z, F (z) = λz + ε(z)z where the constant λ 0 and ε(z) 0 as z, then F (z) = 0 exactly in one point z = z 0 C. Proof: The result a) is essentially from [28]. For the convenience of the reader we will outline a proof for it after first proving b): The continuity of F (z) = λz + ε(z)z and an application of the degree theory [29] or an appropriate homotopy argument show that F is surjective and consequently there exists at least one point z 0 C such that F (z 0 ) = 0. To show that F can not have more zeros, let z 1 C and choose a large disk B = B(0, R) containing both z and z 0. If R is so large that ε(z) < λ/2 for z = R, then F { z =R} is homotopic to identity relative to C \ {0}. Next we express (3.9) in the form (3.10) F = ν(z) F + A(z)F where ν(z) κ < 1 and A(z) γ(z) for almost every z C. Now Aχ B L r (C) for all 1 r p = min{p, 1 + 1/κ} and we obtain from Theorem 3.2 that (I νs) 1 (Aχ B ) L r for all p /(p 1) < r < p. 13

14 Next we define, cf. [28], η = P ( (I νs) 1 (Aχ B ) ). By Proposition 3.1, η C 0 (C) and clearly we have (3.11) η ν η = A(z), z B. By a differentation we see that the function (3.12) g = e η F satisfies (3.13) g ν g = 0, z B. Since η W 1,r (C) by Proposition 3.1, also g W 1,r loc (C) and thus g is quasiregular in B. As such, see e.g. [12] Theorem 1.1.1, g = h ψ, where ψ : B B is a quasiconformal homeomorphism and h holomorphic, both continuous up to a boundary. Since η is continuous, (3.12) shows that g z =R is homotopic to identity relative to C \ {0}, and so is the holomorphic function h { z =R}. Therefore, h has by the principle of the argument ([22], Theorems V.7.1 and VIII.3.5) precisely one zero in B = B(0, R). As already h(ψ(z 0 )) = e η(z 0) F (z 0 ) = 0, there can be no further zeros for F either. This finishes the proof of b). For the claim a) the condition F (z) = ε(z)z is too weak to guarantee F 0 in general. But if γ has a compact support we may choose supp γ B(0, R) and thus the function η solves (3.11) for all z C. Consequently (3.13) holds in the whole plane and g in (3.11) is quasiregular in C. But since F and η are bounded, also g is bounded and thus constant by Liouville s theorem. Now (3.12) gives (3.14) F = C 1 e η, η C 0 (C). With the assumption F (z) 0 as z we then obtain C 1 = 0. We also have the following useful Corollary 3.4 Suppose F W 1,p loc (C) L (C), p > 2, 0 κ < 1 and that γ L p (C) has compact support. If F (z) κ F (z) + γ(z) F (z), z C, then where C 1 is constant and η C 0 (C). F (z) = C 1 e η Proof: This is a reformulation of (3.14) from above. 14

15 4 Complex geometric optics solutions In this section we establish the existence of the solution to (1.4) of the form (4.1) f µ (z, k) = e ikz M µ (z, k) where (4.2) M µ (z, k) 1 = O ( ) 1 z as z. Moreover, it is demonstrated that ( ) Mµ (z, k) (4.3) Re > 0, for all z, k C. M µ (z, k) The importance of (4.3) lies e.g. in the fact that (4.4) ν z (k) = e k (z) M µ(z, k) M µ (z, k) M µ (z, k) + M µ (z, k) appears as the coefficient in a Beltrami equation in the k-variable in Section 8. The result (4.3) clearly implies (4.5) ν z (k) < 1 for all z, k C. We start with Proposition 4.1 Assume that 2 < p < 1 + 1/κ, that α L (C) with supp(α) D and that ν(z) κχ D (z) for almost every z D. Define the operator K : L p (C) L p (C) by Kg = P ( I νs ) 1 (αg). Then K : L p (C) W 1,p (C) and I K is invertible in L p (C). Proof: First we note that by Theorem 3.2, I νs is invertible in L p and by Proposition 3.1 (iii) the operator K : L p (C) L p (C) is well defined and compact. Note that supp ( (1 νs) 1 αg ) D. Thus, by Fredholm s alternative, we need to show that I K is injective. So suppose that g L p (C) satisfies ( (I ) 1 ) (4.6) g = P νs (αg). 15

16 By Proposition 3.1 (ii) g W 1,p and thus by (4.6) or equivalently g = ( I νs ) 1 (αg) (4.7) g ν g = αg. Finally, from (4.7) it follows that g is analytic outside the unit disk. This together with g L p (C) implies ( ) 1 g(z) = O for z. z Thus the assumptions of Proposition 3.3 a) are fulfilled and we must have g 0. It is not difficult to find examples showing that Proposition 4.1 fails for p 2 and for p 1 + 1/κ. We are now ready to establish the existence of the complex geometrical optics solutions to (1.4). Theorem 4.2 For each k C and for each 2 < p < 1 + 1/κ the equation (1.4) admits a unique solution f W 1,p loc (C) of the form (1.5) such that the asymptotic formula (1.6) holds true. In particular, f(z, 0) 1. Proof: If we write and plug this to (1.4) we obtain f µ (z, k) = e ikz M µ (z, k) = e ikz (1 + ω(z)) (4.8) ω e k µ ω = αω + α where e k is defined in (1.12) and (4.9) α(z) = ike k (z)µ(z). Hence (4.10) ω = ( I e k µs ) 1 (αω + α). If now K is defined as in Proposition 4.1 with ν = e k µ we get (4.11) ω Kω = K(χ D ) L p (C). 16

17 Since by Proposition 4.1 the operator I K is invertible in L p (C), and by (4.8) ω is analytic in C \ D the claims follow by (4.10) and (4.11). Next, let f µ (z, k) = e ikz M µ (z, k) and f µ (z, k) = e ikz M µ (z, k) be the solutions of Theorem 4.2 corresponding to conductivities σ and σ 1, respectively. Proposition 4.3 For all k, z C we have ( ) Mµ (z, k) (4.12) Re > 0. M µ (z, k) Proof: Firstly, note that (1.4) implies for M ±µ (4.13) M ±µ µe k M ±µ = ikµe k M ±µ. Thus we may apply Corollary 3.4 to get (4.14) M ±µ (z) = exp(η ± (z)) 0 and consequently M µ /M µ is well defined. Secondly, if (4.12) is not true the continuity of M ±µ and lim z M ±µ (z, k) = 1 imply the existence of z 0 C such that M µ (z 0, k) = itm µ (z 0, k) for some t R \ {0}. But then g = M µ itm µ satisfies g = µ (e k g), ( ) 1 g(z) = 1 it + O, as z. z According to Corollary 3.4 this implies g(z) = (1 it) exp(η(z)) 0, contradicting the assumption g(z 0 ) = 0. 5 k -equations We will prove in this section the k -equation (1.8) for the complex geometrical optics solutions. We begin by writing (1.4)-(1.6) in the form (5.1) M µ = µ (e k M µ ), M µ 1 W 1,p (C). 17

18 By introducing a R-linear operator L µ, we see that (5.1) is equivalent to L µ g = P ( µ (e k g) ) (5.2) (I L µ )M µ = 1. The following refinement of Proposition 4.1 will serve as the main tool in proving (1.8). Below we will study functions of the form f = constant + f 0, where f 0 W 1,p (C). The corresponding Banach space is denoted by W 1,p (C) C. Theorem 5.1 Assume that k C and µ L comp(c) with µ κ < 1. Then for 2 < p < 1 + 1/κ the operator is bounded and invertible. Proof: We write L µ (g) as I L µ : W 1,p (C) C W 1,p (C) C (5.3) L µ (g) = P ( µe k g ikµe k g ) Proposition 3.1 (ii) now yields that (5.4) L µ : W 1,p (C) C W 1,p (C) is bounded. Thus we need to show that I L µ is bijective on W 1,p (C) C. To this end assume (5.5) (I L µ )(g + C 0 ) = h + C 1 for g, h W 1,p (C) and for constants C 0, C 1. This yields C 0 C 1 = g h L µ (g + C 0 ) which by (5.4) gives C 0 = C 1. By differentiating, rearranging and by using the operator K µ from Proposition 4.1 with α = ikµe k and ν = µe k we see that (5.5) is equivalent to (5.6) g K µ (g) = K µ (C 0 χ D ) + P [ (1 µe k S) 1 h ]. Since the right hand side belongs to L p (C) for each h W 1,p (C) this equation has a unique solution g W 1,p (C) by Proposition 4.1. As an immediate corollary we get the following important 18

19 Corollary 5.2 The operator I L 2 µ is invertible on W 1,p (C) C. Proof: Since L µ = L µ we have I L 2 µ = (I L µ )(I L µ ). Next, we make use of the differentiability properties of the operator L µ. For later purposes it will be better to work with L 2 µ which can be written in the following convenient form (5.7) L 2 µg = P ( µ ( + ik) 1 µ( + ik)g ) where the operator ( + ik) 1 is defined by (5.8) ( + ik) 1 g = e k 1 (e k g). Note that many mapping properties of this operator follow from Proposition 3.1. Moreover, we have Lemma 5.3 Let p > 2. Then the operator valued map k ( + ik) 1 is continuously differentiable in C, in the uniform operator topology: L p (D) W 1,p loc (C). Proof: The lemma is a straightforward reformulation of [19], Lemma 2.2, where slightly different function spaces were used. Note that W 1,p loc (C) has the topology given by the seminorm f n = f W 1,p (B(0,n)), n N. Combining Lemma 5.3 with (5.7) shows that k L 2 µ is a C 1 - family of operators L 2 µ : W 1,p (C) C W 1,p (C) C in the uniform operator topology. If we iterate the equation (5.2) once we get (5.9) M µ = 1 + P (µ e k ) + L 2 µ(m µ ). Therefore the above lemma shows that k M µ (z, k) is a continuously differentiable family of functions in W 1,p (C) C, p > 2. In particular, for each fixed z C, M µ (z, k) is continuously differentiable in k. An alternative way to see this is to note that k L µ is smooth in the operator norm topology of L (W 1,p (C) C) and then use Theorem 5.1. This gives by (5.2) that for fixed z the map k M µ (z, k) is, indeed, C -smooth. Furthermore, with respect to the first variable M µ (z, k) is complex analytic in C \ D by (5.1), with development (5.10) M µ (z, k) = 1 + b n (k)z n, for z > 1. n=1 We define the scattering amplitude corresponding to M µ to be (5.11) t µ (k) = b 1 (k). 19

20 Equation (5.1) implies, as µ is real valued, that (5.12) t µ (k) = 1 µ (e k M µ ) dm. π D Beals and Coifman [6] introduced the idea of studying the k-dependence of operators associated to complex geometric optics solutions. We will use the Beals-Coifman principle in the following form: Lemma 5.4 Suppose g W 1,p (C) C is fixed. Then (5.13) k ( e k 1 µ e k g ) = it µ (g; k)e k where (5.14) t µ (g; k) = 1 π µ (e k g) dm. Proof: For f L p comp(c) we have (e k 1 e k f)(z) = 1 π C C e k (ξ z) f(ξ) dm(ξ) ξ z Using this representation [19], Lemma 2.2, shows that (5.15) k (e k 1 e k f)(z) = k ( ( + ik) 1 f ) (z) = i f(k)e k (z) where (5.16) f(k) = 1 π C e k (ξ)f(ξ) dm(ξ). ( By rewriting the left hand side of (5.13) in the form k ( + ik) 1 µ( + ik)g ) we see that the claim follows from (5.15). To get rid of the second term on the right hand side of (5.9) we introduce (5.17) (5.18) In particular, (5.9) gives F + = 1 2 (M µ + M µ ), F = ie k 2 ( Mµ M µ ). (5.19) F + = 1 + L 2 µf +. 20

21 From Lemma 5.4 and (5.7) one has k L 2 µ(g) = it µ (g; k)p (µ e k ) for every g W 1,p (C) C. Hence a differentiation of (5.19) yields (5.20) ( I L 2 µ ) ( k F + ) = iτ µ (k)p (µ e k ) where the scattering coefficient τ µ (k) is (5.21) τ µ (k) t µ (F + ; k) = 1 2 (t µ(k) t µ (k)). Note that this is consistent with (1.9). One way to identify k F + is by readily observing that the unique solution of (5.20) has also other realizations. Namely, if one subtracts the equation (5.9) applied to M µ from the same equation applied to M µ, one obtains after using L 2 µ = L 2 µ that (5.22) (I L 2 µ)(e k F ) = ip (µ e k ). Thus by Corollary 5.2, (5.20) and (5.22) we have proven the first part of Theorem 5.5 For each fixed z C, the functions k F ± (z, k) are continuously differentiable with a) k F + (z, k) = τ µ (k)e k (z)f (z, k), b) k F (z, k) = τ µ (k)e k (z)f + (z, k). Proof: The differentiability is clear since M ±µ (z, k) are continuously differentiable in k. Hence we are left proving b). We start by adding and subtracting (5.1) for M µ and M µ to arrive at the equations (5.23) (5.24) F + = 1 i 1 µ F, F = ie k 1 µ e k F +. By differentiating the second equation with respect to k and by applying Lemma 5.4 we get (5.25) k F = τ µ (k)e k + ie k 1 µ (e k k F + ). Combining this with part a) we have (5.26) k F = τ µ (k)e k (1 + i 1 µ F ). 21

22 This together with (5.23) yields b). We close this section by returning to the functions (5.27) h + = 1 2 (f µ + f µ ) = e ikz F + and (5.28) h = i 2 (f µ f µ ) = e ikz F. These expressions with Theorem 5.5 give immediately the identities (1.8). Note that by Theorem 5.5, k h ± (z, k) is C 1 in C, for each fixed z. 6 From Λ σ to τ We next prove that the Dirichlet to Neumann operator Λ σ uniquely determines f µ (z) and f µ (z) at the points z that lie outside D and moreover that Λ σ determines τ µ (k) for all k C. Proposition 6.1 If σ and σ are two conductivities satisfying the assumptions of Theorem 1, then if µ and µ are the corresponding Beltrami coefficients, we have (6.1) f µ (z) = f µ (z) and f µ (z) = f µ (z) for all z C \ D. Proof: We assume Λ σ = Λ σ which by Proposition 2.3 implies that H µ = H µ. Since Λ σ by the same proposition determines Λ σ 1 it is enough to prove the first claim of (6.1). From (2.9) we see firstly that the projections Q µ = Q µ and thus by Lemma 2.2 Q µ (f f) = constant where we have written f = f µ D and f = f µ D. By using Lemma 2.2 again we see that there exists a function G H 1 (D) such that G satisfies (2.1) in D and G D = f f. Define then G outside D by G(z) = f µ (z) f µ (z), z 1, 22

23 to get a global solution to (6.2) G(z) µ(z) G(z) = 0, z C. Since G is a Hloc 1 -solution to (6.2), the general smoothness properties of quasiregular mappings [3] give G W 1,p loc (C) for all 2 p < 2 + 1/κ. This regularity can also be seen readily from Theorem 3.2, since the compactly supported function h = f µ f µ G satisfies h = (1 µs) 1 ( χ D f µ µ f µ ). Finally, from the above we obtain that the function G 0, defined by G 0 (z) = e ikz G(z), belongs to W 1,p (C) and satisfies G 0 (z) = O(1/z) with G 0 e k µ G 0 = ikµg 0. By Proposition 3.3 a) the function G 0 must hence vanish identically, which proves (6.1). Corollary 6.2 The operator Λ σ uniquely determines t µ, t µ and τ µ. Proof: The claim follows immediately from Proposition 6.1, (1.5) and from the definitions (5.11) of the scattering coefficients. From the results of Section 5 it follows that the coefficient τ µ is continuously differentiable in k and vanishes at the origin. However, the global properties of τ µ need a different approach. To this end we use a simple application of the Schwarz s lemma. Proposition 6.3 The complex geometric optics solutions f ±µ (z, k) = e ikz M ±µ (z, k) satisfy for z > 1 and for all k C (6.3) M µ (z, k) M µ (z, k) M µ (z, k) + M µ (z, k) 1 z. Moreover, for the scattering coefficient τ µ (k) we have (6.4) τ µ (k) 1 for all k C. 23

24 Proof: Fix the parameter k C and denote m(z) = M µ(z, k) M µ (z, k) M µ (z, k) + M µ (z, k). Then by Proposition 4.3, m(z) < 1 for all z C. Moreover, m is C-analytic in z C\D, m( ) = 0, and thus by Schwarz s lemma we have m(z) 1/ z for all z C\D. Since (5.11) and (5.21) give lim z zm(z) = τ µ, both claims of the proposition follow. 7 Subexponential growth We know from Section 4 and (4.1), (4.14) that the complex geometric optics solution f µ from (1.4) can be written in the exponential form. Here we begin by a more detailed analysis of this fact. For later purposes we also need to generalize the situation a bit by considering complex Beltrami coefficients µ λ of the form µ λ = λµ, where the constant λ D and µ is as before. Precisely as in Section 4 one can show the existence and uniqueness of f λµ W 1,p loc (C) satisfying (7.1) f λµ = λµ f λµ and (7.2) f λµ (z, k) = e ikz (1 + O ( )) 1 z as z. Lemma 7.1 The function f λµ admits a representation (7.3) f λµ (z, k) = e ikϕ λ(z,k), where for each fixed k C \ {0} and λ D, the function ϕ λ (, k) : C C is a quasiconformal homeomorphism that satisfies ( ) 1 (7.4) ϕ λ (z, k) = z + O for z z and (7.5) ϕ λ (z, k) = k k µ λ(z) (e k ϕ λ )(z, k) ϕ λ (z, k), z C. 24

25 Proof: Since the argument k is fixed we drop it from the notations and write simply f λµ (z, k) = f λµ (z), ϕ λ (z, k) = ϕ λ (z), etc. Denote Then (1.4) gives µ 1 (z) = µ λ (z) f µ(z) f µ (z). (7.6) f λµ = µ 1 f λµ. On the other hand, by the general theory of quasiconformal maps ([2], [12], [18]) there exists a unique quasiconformal homeomorphism ϕ λ H 1 loc (C) satisfying (7.7) ϕ λ = µ 1 ϕ λ and having the asymptotics (7.8) ϕ λ (z) = z + O ( ) 1 z as z. Moreover, any H 1 loc -solution to (7.6) is obtained from ϕ λ by post-composing with an analytic function ([12], Theorem ). In particular, f λµ (z) = h ϕ λ (z) where h : C C is an entire analytic function. But h ϕ λ (z) exp(ikϕ λ (z)) = f λµ (z) exp(ikϕ λ (z)) has by (1.5), (1.6) and (7.8) the limit 1 as the variable z. Thus h(z) e ikz. Finally, (7.5) follows immediately from (1.4) and (7.3). Note that the results in Section 4 show that (7.4), (7.5) has a unique solution. The existence of such a solution can also be directly verified by using Schauder s fixed point theorem [13], [5]. The result of Lemma 7.1 demonstrates that after a change of coordinates z ϕ(z) the complex geometric optics solution f µ is simply an exponential function. The main goal of this section is to show 25

26 Theorem 7.2 If ϕ λ satisfies (7.4) and (7.5) then ϕ λ (z, k) z uniformly in z C and λ D, as k. We have splitted the proof of Theorem 7.2 to several lemmata. Lemma 7.3 Suppose ε > 0 is given. Suppose also that for µ λ (z) = λµ(z) we have (7.9) f n = µ λ S n µ λ S n 1 µ λ µ λ S 1 µ λ where S j : L 2 (C) L 2 (C) are Fourier multiplier operators, each with a unimodular symbol. Then there is a number R n = R n (κ, ε) depending only on µ, n and ε such that (7.10) f n (ξ) < ε for ξ > R n. Proof: Clearly it is enough to prove the claim for λ = 1. Recall that for the Fourier transform φ we use the definition (5.16). By assumption Ŝ j g(ξ) = m j (ξ)ĝ(ξ) where m j (ξ) = 1 for ξ C. We have by (7.9) (7.11) f n L 2 µ n L µ L 2 πκ n+1 since supp(µ) D. Choose first ρ n so that (7.12) µ(ξ) 2 dm(ξ) < ε 2. ξ >ρ n After this choose ρ n 1, ρ n 2,..., ρ 1 inductively so that for l = n 1,..., 1 (7.13) π Finally, choose ρ 0 so that ξ >ρ l µ(ξ) 2 dm(ξ) ε 2 ( n j=l+1 πρ j ) 2 ( n ) 1 (7.14) µ(ξ) < επ n ρ j, when ξ > ρ 0. j=1 26.

27 All these choices are possible since µ L 1 L 2. Now, we set R n = n j=0 ρ j and claim that (7.10) holds for this choice of R n. Hence assume that ξ > n j=0 ρ j. We have (7.15) f n (ξ) µ(ξ η) f n 1(η) dm(η) ξ η ρ n + µ(ξ η) f n 1(η) dm(η). ξ η ρ n But if ξ η ρ n then η > n 1 j=0 ρ j. Thus, if we denote { } n = sup f n n (ξ) : ξ > ρ j it follows from (7.15) and (7.11) that n n 1 (πρ 2 n) 1/2 µ L 2 + µ(η) 2 dm(η) j=0 1/2 f n 1 L 2 η ρ n πρ n κ n 1 + κ n π µ(η) 2 dm(η) 1/2 η ρ n for n 2. Moreover, the same argument shows that 1 πρ 1 κ sup{ µ(ξ) : ξ > ρ 0 } + κ π µ(η) 2 dm(η) η >ρ 1 In conclusion, after an iteration we have ( n ) n (κπ) n ρ j sup{ µ(ξ) : ξ > ρ 0 } j=1 ( n n + κ n l=1 j=l+1 ) πρ j π η >ρ l 1/2 µ(η) 2 dm(η). 1/2. 27

28 With the choices (7.12) (7.14) this leads to which proves the claim. n (n + 1)κ n ε ε 1 κ, Our next goal is to use Lemma 7.3 to obtain a similar asymptotic result as in Theorem 7.2 for a solution to a linear equation somewhat similar to (7.5). Proposition 7.4 Suppose ψ Hloc 1 (C) satisfies (7.16) ψ = λ k k µ(z)e k(z) ψ, and ( ) 1 ψ(z) = z + O as z. z Then ψ(z, k) z, uniformly in z C and λ D, as k. For Proposition 7.4 we need some preparations. First, as S L p L p 1 when p 2, we can choose a δ κ > 0 so that κ S L p L p < 1 whenever 2 δ κ p 2 + δ κ. With this notation we then have Lemma 7.5 Let ψ be the solution of (7.16) and let ε > 0. Then one can decompose ψ in the following way: ψ = g + h where (i) h(, k) L p < ε for 2 δ κ p 2 + δ κ, uniformly in k, (ii) g(, k) L p C 0 = C 0 (κ), uniformly in k and (iii) ĝ(ξ, k) 0 as k, where in (iii) convergence is uniform on compact subsets of the ξ-plane and also uniform in λ D. The Fourier transform is with respect to the first variable only. Proof: We may solve (7.16) by Born-series which converge in L p, ψ = n=0 ( k ) n ( k λµe k(z)s k ) k λµe k. Let h = n=n 0 ( k ) n ( k λµe k(z)s k ) k λµe k. 28

29 Then h L p π 1/p κn 0+1 S n 0 L p L p 1 κ S L p L p. We obtain (i) by choosing n 0 large enough. The remaining part clearly satisfies (ii) with a constant C 0 that is independent of k and λ. To prove (iii) we first note that S(e k φ) = e k S k φ where (S k φ) (ξ) = m(ξ k) φ(ξ) and m(ξ) = ξ/ξ. Consequently, (µe k S) n µe k = e (n+1)k µs nk µs (n 1)k µs k µ and so Therefore n 0 ( g = k ) j k λ e jk µs (j 1)k µ µs k µ. j=1 n 0 g = e jk G j j=1 where by Lemma 7.3, Ĝj(ξ) < ε whenever ξ > R = max j n0 R j. As (e jk G j ) (ξ) = Ĝj(ξ + jk), for any fixed compact set K 0 we can take k so large that jk + K 0 C \ B(0, R) for each 1 j n 0. Then This proves (iii). sup ĝ(ξ, k) n 0 ε. ξ K 0 Proof of Proposition 7.4: We show first that when k, ψ 0 weakly in L p, 2 δ κ p 2 + δ κ. For this suppose f 0 L q, q = p/(p 1), is fixed and choose ε > 0. Then there exists f C0 (C) such that f 0 f L q < ε and so by Lemma 7.5 f 0, ψ εc 1 + f(ξ)ĝ(ξ, k) dm(ξ). Choose first R so large that f(ξ) 2 dm(ξ) ε 2 C\B(0,R) 29

30 and then k so large that ĝ(ξ, k) ε/( πr) for all ξ B(0, R). Now f(ξ)ĝ(ξ, k) dm(ξ) f(ξ)ĝ(ξ, k) dm(ξ) + f(ξ)ĝ(ξ, k) dm(ξ) B(0,R) ε( f L 2 + g L 2) C 2 (f)ε. C\B(0,R) The bound is the same for all λ, hence sup λ D f 0, ψ 0 as k. To prove the uniform convergence of ψ itself we write (7.17) ψ(z, k) = z 1 1 ψ(ξ, k) dm(ξ). π ξ z D Here note that supp( ψ) D and χ D (ξ)/(ξ z) L q for all q < 2. Thus by the weak convergence we get (7.18) ψ(z, k) z as k, for each fixed z C, but uniformly in λ D. On the other hand as sup k ψ L p C 0 (κ) <, for all z sufficiently large ψ(z, k) z < ε, uniformly in k C and λ D. Moreover, (7.17) shows also that the family {ψ(, k) : k C, λ D} is equicontinuous. Combining all these observations shows that the convergence in (7.18) is uniform in z C and λ D. Finally we proceed to the nonlinear case: So assume that ϕ λ satisfies (7.4) and (7.5). Since ϕ is a (quasiconformal) homeomorphism we may consider its inverse ψ λ : C C, (7.19) ψ λ ϕ λ (z) = z, which also is quasiconformal. By differentiating (7.19) with respect to z and z one obtains that ψ satisfies (7.20) (7.21) ψ λ = k k λ(µ ψ λ)e k ψ λ and ( ) 1 ψ λ (z, k) = z + O as z. z Proof of Theorem 7.2: It is enough to show that (7.22) ψ λ (z, k) z 30

31 uniformly in z and λ as k. To prove this we need to recall some further facts from quasiconformal mappings. Let us use the notation (7.23) ( ) 1 Σ κ = {g Hloc(C) 1 : g = ν g, ν κχ 4D and g = z + O as z }. z Note that with the above normalization at all elements g Σ κ are homeomorphisms [2]. Also, the use of χ 4D will become clear soon. Lemma 7.6 a) The family Σ κ is compact in the topology of uniform convergence on C. b) Suppose that f, g Σ κ, 1 + κ < p < 1 + 1/κ and that ε > 0 is so small that (1 + ε)p < 1 + 1/κ. Then f g p dm C(p, ε) ν f ν g p 1+ε ε dm ε 1+ε C C where ν f = f f and ν g = g g. Proof: The claim a) follows from [18], Theorem II.51. Furthermore, since g = 1 + S( g) for each g Σ κ, we have f g = (I ν f S) 1 ( ν f ν g + (ν f ν g )S( g) ), f, g Σ κ. Applying Theorem 3.2 and Hölder s inequality gives then the claim b). The support of µ ψ λ need not anymore be contained in D. However, by Koebe s 1/4-theorem, see e.g. [1] Corollary 5.3, ϕ λ (D) 4D and thus supp(µ ψ λ ) 4D. Therefore by Lemma 7.6 a) we have sequences k n and sequences λ n λ D such that ψ λn (, k n ) ψ uniformly, with ψ Σ κ. To prove Theorem 7.2 it is enough to show that for any such sequence ψ (z) z. Hence assume that we have such a limit function ψ. We then consider the Hloc 1 -solution Φ(z) = Φ λ(z, k) of Φ = k k λ(µ ψ )e k Φ, ( ) 1 Φ(z) = z + O as z. z 31

32 This is now a linear Beltrami equation which [2] has a unique solution Φ Σ κ for each k C. According to Proposition 7.4 (7.24) Φ λ (z, k) z as k. Secondly, when 2 < p < 1 + 1/κ, by Lemma 7.6 b) (7.25) ψ λn (z, k n ) Φ λ (z, k n ) = 1 π 4D 1 ξ z ( ψ λn (ξ, k n ) Φ λ (ξ, k n ) ) dm(ξ) C 1 ( ψ λn (ξ, k n ) Φ λ (ξ, k n ) ) L p ( C 2 λ n λ + C 2 µ ψλn (ξ, k n ) ) µ ( ψ (ξ) ) p(1+ε) ε dm(ξ) 4D ε p(1+ε) Lastly, we use the fact [3] that for all 2 < p < 1+1/κ and for all g = ψ 1, ψ Σ κ, we have for the Jacobian J g that (7.26) J g (z) p/2 dm(z) g p dm C(κ) <.. D D where C(κ) depends only on κ. Again, the bound can be deduced also from Theorem 3.2 since g = (I ν g S) 1 ν g and g = 1 + S( g). We use this estimate in the cases ψ(z) = ψ λn (z, k n ) and ψ = ψ. Namely, we have for each η C0 (D) that µ(ψ) η(ψ) p(1+ε) ε = µ η p(1+ε) ε J g (7.27) 4D D ( D µ η p 2 (1+ε) ε(p 2) ) (p 2)/p ( ) 2/p. Jg p/2 D Since µ can be approximated in the mean by smooth η, the last term in (7.27) can be made arbitrarily small. Since by uniform convergence η(ψ λn (z, k n )) η(ψ (z)) we see that the last bound in (7.25) converges to zero as λ n λ and k n k. In view of (7.24) and (7.25) we have established that ψ λn (z, k n ) z and that ψ (z) z. The theorem is proved. 32

33 8 Transport matrix The gradient of a quasiregular map can vanish only on a set of Lebesque measure zero ([2], p. 34). By the equation (1.4) the map f µ can vanish only on points z where the whole gradient of f µ vanishes. This means that we can recover µ and hence σ from f µ by the formulae (8.1) µ = ( f µ ) 1 f µ and σ = 1 µ 1 + µ. Next, let u 1 = h + ih and u 2 = i(h + + ih ) be defined as in (1.14). From Lemma 2.1 and (1.7) we see that u 1, u 2 satisfy the conductivity equations (1.15). With respect to the parameter k the u j s are C 1 -mappings, c.f. Theorem 5.5. A straight forward derivation using (1.8) shows also that at each point z C, both u 1 (z, k) and u 2 (z, k) satisfy the k -equation (8.2) k u(z, k) = i τ µ(k)u(z, k). It is clear that the pair {u 1 (z, k), u 2 (z, k)} determines the pair {f µ (z, k), f µ (z, k)} and vice versa. Thus by Proposition 6.1 the Dirichlet to Neumann map Λ σ uniquely determines the functions u 1 and u 2 outside D. To prove Theorem 1 it therefore suffices to transport these solutions from outside to inside D by using the data Λ σ. For this purpose we will employ (8.2) and the fact from Corollary 6.2 that Λ σ uniquely determines the scattering coefficient τ µ. We have only shown that τ µ is bounded and that u j (z, k)e ikz behaves subexponentially as k, and these facts alone are much too weak to guarantee the uniqueness of solutions to the pseudoanalytic equation (8.2). To remedy this we need to understand the transport matrix from (1.18). We start by arguing that u 1 and u 2 are R-linearly independent, i.e. that (8.3) u 2 (z, k) 0 and u 1 (z, k)/u 2 (z, k) R holds for all z, k C. By Proposition 4.3 we have h (z, k) h + (z, k) < 1 and h +(z, k) 0 for all z, k C. Since u 1 u 2 = i h + ih h + + ih this proves (8.3) and enables us to define the transport matrix T σ z,z 0 (k) as in (1.18). It also shows that T σ z,z 0 (k) is invertible. 33

34 It was discovered by Bers [7] that the coefficients connecting different solutions to the same pseudoanalytic equation as in (1.17) give rise to a quasiregular mapping. In [7] these mappings are called pseudoanalytic functions of the second kind. In our case this means that differentiating u 1 (z, k) = a 1 (z, z 0 ; k)u 1 (z 0, k) + a 2 (z, z 0 ; k)u 2 (z 0, k) with respect to k and using (8.2) with (1.14) gives for α = a 1 + ia 2 (8.4) k α(z, z 0 ; k) = ν z0 (k) k α(z, z 0 ; k), ν z0 (k) = i h (z 0, k) h + (z 0, k). Note that α(z, z 0 ; k) inherits the continuous differentiability with respect to k from the u j s. For an explicit expression for α in terms of u 1 and h ± see (8.8) below. Moreover, the second row of T σ z,z 0 (k) gives similarly a solution β = b 1 +ib 2 for the same equation (8.4). In fact, if we write the µ dependence explicitely for u i : u i = u i (µ), i = 1, 2, then (8.5) u 2 (µ) = iu 1 ( µ) and u 1 (µ) = iu 2 ( µ). Thus iβ µ = α µ. We have now shown that the rows of the transport matrix produce quasiregular mappings, with respect to k, satisfying (8.4). Our next task is to determine their asymptotic behaviour at. Proposition 8.1 Suppose z 0 C, z 0 > 1. Then: a) For each fixed k 0 and z 0 we have with respect to z where v L. α(z, z 0 ; k) = exp(ikz + v(z)) b) For each fixed z and z 0 we have with respect to k that (8.6) α(z, z 0 ; k) = exp(ik(z z 0 ) + kε(k)) where ε(k) 0 as k. Up to the factor i, β has the same asymptotics. Proof: The last claim is clear from (8.5). For a) we write (8.7) u 1 = 1 ( ) fµ + f µ + f µ f µ 2 ( = f µ 1 + f ) 1 ( µ f µ 1 + f ) µ f µ. f µ + f µ f µ + f µ 34

35 All factors in the product are nonvanishing. Taking the logarithm and applying (1.5), (1.6) leads to ( ( )) 1 u 1 (z, k) = exp ikz + O k. z On the other hand, dropping temporarily the fixed k from the notations, or u 1 (z) = αh + (z 0 ) iαh (z 0 ) u 1 (z) h + (z 0 ) = α iαh (z 0 ) h + (z 0 ). This gives ( (8.8) α = 1 h (z 0 ) ) 2 1 ( ) h + (z 0 ) 1 + i h (z 0 ) u 1 (z) u 1 (z) h + (z 0 ) u 1 (z) h + (z 0 ). According to Proposition 6.3, h (z 0 )/h + (z 0 ) 1/ z 0 < 1 while h + (z 0 ) is constant for fixed k and z 0. This proves a). To prove b) note that h + = ( 1 + i h ) 1 ( f µ and u 1 = h + 1 i h ). h + h + Again, the factors are continuous and point wise non-vanishing. Therefore the identity (8.8) and Theorem 7.2 reduce the proof of b) to Lemma 8.2 For each fixed z C, h (z, k) h + (z, k) 1 e k ε(k). Proof: By the definition of h + and h it suffices to show (8.9) inf f µ f µ t + e it f µ + f µ e k ε(k). For this, define Φ t = e it/2 (f µ cos t/2 + if µ sin t/2). Then for each fixed k, ( ( )) 1 Φ t (z, k) = e ikz 1 + O k z as z 35

36 and Φ t = µe it Φ t. Thus Φ t = f λµ, λ = e it, the exponentially growing solution of (7.1) and (7.2) from Section 7. But (8.10) f µ f µ f µ + f µ + e it = 2eit Φ t f µ + f µ = f λµ f µ 2e it 1 + M µ /M µ. By Theorem 7.2 (8.11) e k ε 1(k) M ±µ (z, k) e k ε 1(k) and (8.12) e k ε2(k) inf f λµ (z, k) λ D f µ (z, k) sup f λµ (z, k) f µ (z, k) e k ε 2(k) λ D where ε j (k) 0 as k. Since Re(M µ /M µ ) > 0 the inequality (8.9) follows from (8.10), (8.11) and (8.12). This finishes the proof of Lemma 8.2 and thus also the proof of Proposition 8.1. Now the last remaining obstacle is Theorem 2; that is, we need to prove that the data determines the transport matrices Tz,z σ 0 (k). We know that in the k-variable the rows α and β of Tz,z σ 0 are quasiregular mappings satisfying (8.4) and having the asymptotics given by Proposition 8.1 b). It is not clear if the asymptotics (8.6) are strong enough to determine the individual solution. However, if we consider the entire family {Tz,z σ 0 : z C}, then the uniqueness does hold: Proof of Theorem 2: Let z 0 > 1 and k 0. Let α µ (z, z 0, k) and α µ (z, z 0, k) be defined by (1.17) and (1.19). Since neither of α µ, α µ vanishes at any point we can define the corresponding logarithms δ µ and δ µ by (8.13) δ µ (z, z 0 ; k) = log α µ (z, z 0, k) = ik(z z 0 ) + kε 1 (k) δ µ (z, z 0 ; k) = log α µ (z, z 0, k) = ik(z z 0 ) + kε 2 (k) where for k, ε j (k) 0 by Proposition 8.1 b). Moreover, by Theorem 4.2, δ µ (z, z 0 ; 0) δ µ (z, z 0 ; 0) 0 for all z C, z 0 > 1. In addition, z δ µ (z, z 0 ; k) is continuous, δ µ (z 0, z 0 ; k) = 0 and we have ( (8.14) δ µ (z, z 0 ; k) = ikz 1 + v ) k(z), k 0, ikz 36

37 where by Proposition 8.1 a), v k L (C) for each fixed k C. The use of degree theory or homotopy argument [29] gives that z δ µ (z, z 0 ; k) is also surjective C C. To prove the theorem it suffices to show that if Λ σ = Λ σ, then we have (8.15) δ µ (z, z 0 ; k) δ µ (w, z o ; k), for z w and k 0. Namely then (8.15) and the surjectivity of z δ µ (z, z 0 ; k) show that necessarily we have δ µ (z, z 0, k) = δ µ (z, z 0, k) for all k, z C and z 0 > 1. Hence α µ α µ. By (8.5) we have β µ = β µ as well and hence that T σ z,z 0 (k) T σ z,z 0 (k). To show (8.15) fix z w and note that by (8.4) δ µ and δ µ satisfy (8.16) k δ = ν z0 (k)e δ δ k δ, k C, where by Proposition 6.1 and the assumption Λ σ = Λ σ, the coefficient ν z0 the same for both δ µ and δ µ. The difference is g(k) = δ µ (w, z o ; k) δ µ (z, z 0 ; k) satisfies the equation ( ) k g ν z0 e (δµ δµ) k g = ν z0 k δ µ e (δµ δµ) e (δ µ δ µ ). In other words, there exist functions η and γ such that (8.17) k g η k g = γg with η ν z0 1/ z 0 < 1 and γ 2 ν z0 k δ µ 2 k δ µ. From (8.13) we have g(k) = i(w z)k + kε(k). Since α µ, α µ are C 1 with respect to k, we see that γ is locally bounded with respect to k and we may apply Proposition 3.3 b) (with respect to k) to obtain that g vanishes only at k = 0. This shows (8.15). The proof of Theorem 1 is now immediate. If Λ σ = Λ σ, then by Proposition 6.1 we have u σ j (z) = u σ j (z) for z > 1 and j = 1, 2. Theorem 2 with (1.17) gives then u σ j u σ j and (8.1) that σ σ. Lastly, we describe how our uniqueness proof leads to a constructive procedure for recovering σ. First, by (2.9) and (2.11) the projection operator Q µ can be calculated from Λ σ. Next, by Lemma 2.2 and equation (5.1) the solution M µ (, k) belongs to the space Range(P k µ ) Range(P + ) where the projections P k µ and P + are defined by (8.18) P + (g)(z) = 1 2 (I ih 0)(g)(z) P k µ (g)(z) = e ikz (I Q µ )(e ik g)(z). 37